A particle moves so that its position vector is given by $\overrightarrow{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y},$where $\omega$ is a constant.

Which of the following is true?

Given  $\overrightarrow{r}=\cos \omega t\hat{x}+\sin \omega t\hat{y}$  ____(1)

as we know   $\overrightarrow{\mathrm{v}}=\frac{\mathrm{d}\overrightarrow{\mathrm{r}}}{\mathrm{dt}}=\frac{\mathrm{d}(\cos \omega \widehat{\mathrm{tx}}+\sin \omega \mathrm{ty})}{\mathrm{dt}}$

$\overrightarrow{v}=-\omega \sin \left(\omega t\right)\hat{x}+\omega \cos \left(\omega t\right)\hat{y}-\left(2\right)$

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\frac{d(-\omega \sin (\omega t)\hat{x}+\omega \cos (\omega t\left)\hat{y}\right)}{dt}$

$=-{\omega }^2\cos \left(\omega t\right)\hat{x}-{\omega }^2\sin \left(\omega t\right)\hat{y}$

$\overrightarrow{\mathrm{a}}=-{\omega }^2(\cos (\omega \mathrm{t})\hat{\mathrm{x}}+\sin (\omega \mathrm{t}\left)\hat{\mathrm{y}}\right)-\left(3\right)$

From (1) and (2)
we get

$\begin{array}{c}\overrightarrow{\mathrm{v}}\cdot \overrightarrow{\mathrm{r}}=[-\omega \sin \omega \hat{\mathrm{x}}+\omega \cos \omega \mathrm{ty}].\\ [\cos \omega \mathrm{t}\hat{x}+\sin \omega \hat{\mathrm{y}}]\\ =-\omega \sin \left(\omega t\right)\cos \left(\omega t\right)+\\ \omega \cos \left(\omega t\right)\sin \left(\omega t\right)=0\end{array}$

So we can conclude that  $\overrightarrow{\mathrm{v}} \text{and }\overrightarrow{\mathrm{r}}$ are perpendicular to each other.

From (1) and    $\overrightarrow{\mathrm{a}}=-{\omega }^2\left(\overrightarrow{\mathrm{r}}\right)$

So we can conclude that  $\overrightarrow{a} \text{and }\left(\overrightarrow{r}\right)$ have opposite directions.

$\therefore$  The acceleration is directed towards origin.